Who is the weak* sequential closure of the set of finitely supported measures on the integers?

Let $X$ be a topological space and $Y\subseteq X$, the sequential closure of $Y$ is the set of elements in $X$ that are limit of sequences belonging to $Y$.

Let $\mathcal M_{\text{fin}}(\mathbb Z)$ be the set of finitely supported probability measures on $\mathbb Z$. The general question is: who is the weak* sequential closure of $\mathcal M_{\text{fin}}(\mathbb Z)$? This set should be different from the set of all finitely additive probability measures on $\mathbb Z$, since $L^\infty(\mathbb Z)$ is not separable.

I'm actually interested in a more specific question: does this closure contain at least one (all??) translation invariant finitely additive probability measure?

Valerio

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The finitely support probability measures on $\mathbb Z$ are all members of $\ell^1(\mathbb Z)$. So we could ask a slightly more general question:

What is the sequential closure of (the probability measures in) $\ell^1(\mathbb Z)$ in $\ell^\infty(\mathbb Z)^*$?

If $(a_n)$ is a sequence in $\ell^1(\mathbb Z)$ which converges weak$^*$ in $\ell^\infty(\mathbb Z)^*$, then $f(a_n)$ is a Cauchy sequence for all $f\in \ell^\infty(\mathbb Z)$. Now, $\ell^1(\mathbb Z)$ is weakly sequentially complete (*) and so there is $a\in\ell^1(\mathbb Z)$ with $f(a_n) \rightarrow f(a)$ for each $f\in\ell^\infty(\mathbb Z)$. In other words, the weak$^*$ limit of $(a_n)$ actually lives in $\ell^1(\mathbb Z)$. So the answer is "no" in a strong sense.

(*) All preduals of a von Neumann algebra are weakly sequentially complete (see Takesaki I, Chapter III, Corollary 5.2). There is presumably an easy proof for $\ell^1$, but I don't know a reference off the top of my head.

Edit: The following is prompted by Bill Johnson's comment (and reveals nothing so much as my own ignorance!) A Banach space $E$ is said to have the Schur Property if weakly convergent sequences are norm convergent-- Schur originally showed that $\ell^1$ has the Schur Property. We say that $E$ is weakly sequentially complete (WSC) if whenever $(x_n)$ is weakly Cauchy (meaning that for each $f\in E^*$, the scalar sequence $(f(x_n))$ is Cauchy) then $(x_n)$ is weakly convergent. The Schur property implies WSC-- if $(x_n)$ is weakly Cauchy, then for any increasing sequences $(n_k)$ and $(m_k)$, the sequence $(x_{n_k} - x_{m_k})_k$ is weakly-null, hence norm-null, and hence $(x_n)$ is norm Cauchy. You can find all this in Section 2.3 of Albiac and Kalton's book. For a general measure $\mu$, the space $L^1(\mu)$ is WSC, but doesn't necessarily have the Schur property-- see Section 5.2 of this book.

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In fact, $\ell_1$ has the Schur property. Every weakly Cauchy sequence is norm convergent. You can find this in many basic texts on Banach space theory. – Bill Johnson Jun 29 '11 at 20:02
@Bill-- Yes, I should have added that. What worried me is that e.g. Wikipedia (en.wikipedia.org/wiki/Schur%27s_property) seems to want that we already know $(a_n)$ converges weakly to $a\in \ell^1$ (to then conclude actual norm convergence). Maybe I shouldn't use Wikipedia... – Matthew Daws Jun 29 '11 at 20:13
You can find the proof (of the stronger result Bill alludes to) in e.g. books.google.com/… or books.google.com/… Neither argument is particularly hard. – Matthew Daws Jun 29 '11 at 20:45
Thank you very much! – Valerio Capraro Jun 30 '11 at 7:06

Interesting question. I'd like to know if there is a more elementary answer to it. Here's a candidate start of an answer:

Converging weak* in $l^\infty({\mathbb Z})^*$ implies in particular that $a_n$ is a convergent sequence for all $n$. Let me use the notation $f(a_n)$ to denote the dual pairing between $f \in l^\infty$ and $a_n \in l^1$. From the existence of a weak limit, we see there is a pointwise limit $a = (a_k)$, and hence $f(a_n) \to f(a)$ for all compactly supported $f$. To see that $a$ is in $l^1$, observe that we have the bound $|f(a_n)| \leq C$ for all $||f||_{l^\infty} \leq 1$ and all $n$ because $f(a_n) \to A[f]$ for some bounded, linear functional $A \in (l^\infty)^*$. With this estimate in hand, we can conclude that $f(a_n) \to f(a)$ for all $||f||_{l^\infty} \leq 1$ which vanish at infinity, and therefore $|f(a)| \leq C$ for all such $f$. It then follows that $a \in l^1$ by just the Lebesgue-theoretic definition of an infinite sum of positive numbers as a supremum of finite sums (just choose the $f$ correctly).

What this argument has not shown is that $f(a_n) \to f(a)$ for all $f$ in $l^\infty$. It seems like the only thing that could have gone wrong is if somehow the mass of $|a_n - a|$ could have gone off towards infinity in ${\mathbb Z}$. But if this were the case, it seems like you could cook up an $f \in l^\infty$ so that $f(a_n)$ oscillates, which would contradict the weak convergence of the sequence.

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@Phil: Basically you are starting the proof that $\ell_1$ has the Schur property. If $f_n$ is weakly Cauchy in $\ell_1$ then it weak$^*$ converges ($\ell_1=\co_0^*$) converges to, say, $f$ (which, since $f_n$ must be bounded, is the same as saying that $f_n$ converges coordinatewise to $f$). Subtract off $f$ from each $f_n$ to assume WLOG that $f=0$. If the new $f_n$ does not converge in norm to zero, use a gliding hump argument to get a subsequence with bid disjoint pieces. So far $\ell_1$ could have been any of the familiar sequence spaces. Now use that your space is $\ell_1$. – Bill Johnson Jun 30 '11 at 15:45